Recently I read a comment in an on-line discussion forum regarding the communication range of a handheld radio. An anecdotal report said, in one user's experience, a handheld radio could "barely communicate" over a distance of a half mile. This struck me as being unusual, and it was certainly contrary to my own experience with using handheld VHF radios for many years. That remark prompted me to analyze the communication circuit for two handheld radios to communicate with each other. My findings are below.

Communication Range of Handheld Radio

A common situation in marine radio communication involves use of a handheld radio. This article attempts to analyze the range of communication for such a radio. The transmitter power of the handheld radio is assumed to be 5-Watts. The antenna of the handheld radio is physically shorter than the usual full-size antenna that would be used, and therefore we assume the antenna gain is negative in comparison to a full-size halfwave dipole antenna. Exactly how much less effective the short antenna will be is difficult to assess. Here I will assume the antenna gain is -6 dB compared to isotropic, which would about -8 dB compared to a dipole. There is no transmission line, so the transmission line loss is assumed to be 0 dB. We now analyze the transmitted signal level from a handheld transmitter as described above:

Transmitter power 5-Watt = +37 dBm, where 0 dBm is one milliWatt

Transmission line loss = 0 dB

Antenna gain = -6 dB

Radiated power = +31 dBm

For path loss we will calculate for a distance of 0.5-mile at a frequency of 157-MHz. Assuming free-space attenuation, the path loss will be

Lp = 36.6 + 20log(f) + 20log(d) (f in MHz, d in miles)

For a distance of 0.5-miles this calculates to -74.5 dB. The assumption of a free-space path loss is reasonable if the two antennas are actually in visual sight of each other and there is mostly open water as the intervening terrain.

The signal from the 5-Watt handheld radio after traveling a 0.5-mile path in freespace will be

Transmitter signal + path loss = signal at receive antenna

+31 dBM + -80.5 dB = -49.5 dBm

We assume the handheld receiver sensitivity is as good as a fixed base radio, that is, the receiver is able to produce usable signals from received signals as weak as -113 dBm, or 0.5-microVolts in a 50-Ohm antenna. Now we find the received signal strength at the receiver, including antenna gain and transmission line loss:

Signal level = -49.5 dBm

Antenna gain = -6 dB

Receiver line loss = 0 dB

Received signal = -55.5 dBm

Because the receiver sensitivity is sufficient to work with signals as weak as -113 dBm, and the available signal is much stronger, -55.5 dBm, we have a reserve gain in the path of -55.5 - -113 = 57.5 dB. This means that the signal could fade as much as 57.5 dB and the communication path would still be usable.

The analysis shows that if two 5-Watt VHF Marine Band handheld radio are only 0.5-mile apart and they are in visual sight of each other (which permits a reasonable assumption of the freespace path loss), there should be no difficulty to communicate. The transmitter power, antenna gains, receiver sensitivity, and path loss are such that there is a fade margin of 57.5 dB. This suggests excellent reliability.

If the transmitter power were decreased to 1-Watt, the transmitted signal power would decline to 30 dBm, a reduction of 7 dB. Since the path had a fade margin of 57.5 dB, the reduced transmitter power provides communication with a fade margin of 50.5 dB.

If the antennas of the radios are less effective, let us say only having a gain of -9 dB compared to isotropic, that is a net reduction of only -6 dB in the signal path, still giving a fade margin 51.5 dB.

If we assume both 1-Watt transmitter power reduction and -9 dBi antennas, then the fade margin is still 44.5 dB.

If we assume cross-polarization of the antennas, we can add about -20 dB to the path loss, but we still have a fade margin of 24.5 dB.

If we assume the path loss will be much greater than the free-space assumption, we can deduct the greater path loss from the fade margin calculated. Since there is a fade margin of 57.5-dB, a great deal of added path loss can be tolerated..

These calculations affirm my initial opinion based on experience: two VHF Marine Band handheld radios should be able to communicate over a distance of 0.5-miles with high reliability, as long as the path is a true line-of-sight path, that is, the two radios are in sight of each other. Since we assume that a handheld radio will be held at least 5-feet off the ground, we know the distance to the radio horizon will be

d = (2 x h)^0.5 where d is in miles and h is in feet

d = 3.16-miles

This affirms that the radio horizon of the two radios will overlap.

On the basis of the above calculations, two 157-MHz handheld 5-Watt radios separated by a distance of 0.5-miles and in sight of each other should have no problem communicating. Even if the transmitter power is decreased to 1-Watt, the antenna gain cut in half, and the antennas oriented for cross polarization, there should still be sufficient fade margin in the path to allow good communication.

The one factor which was not varied in the above calculations was the receiver sensitivity. Note that receiver sensitivity can be reduced if the operator of the radio sets the squelch threshold improperly. Local noise interference will also reduce the apparent receiver sensitivity. Another influence on performance will be battery status. Transmitted power will be reduced if the battery voltage is below normal.

Note that if the other station in the communication is using a fixed mount radio with a better antenna, the range improves considerably. The antenna gain of the other station improves both the receiving and transmitting paths of both stations. Since the analysis assume a loss of -6dB for the antennas, substituting an antenna with 3dB gain at the other station improves the fade margin by 9dB on the path in either direction.

For more background see some prior articles on these topics:

Radio Horizon

http://continuouswave.com/radio/radioHorizon.html

Marine VHF Radio Communications

http://continuouswave.com/whaler/reference/VHF.html

Conversion of Receiver Sensitivity

From micro-volts to dBm

http://continuouswave.com/radio/dBm.html

## Communication Range of Handheld Radio

### Re: Communication Range of Handheld Radio

Comparing the range of a handheld radio to a fixed mount radio with a fixed mount antenna

The range of a handheld radio compared to a fixed mount radio and antenna will generally be less. You can estimate the decrease in range by comparing the power output, the antenna gain, and the antenna height.

The power output of a handheld is typically only 5 Watts. A fixed radio can output 25-Watts. A 5-Watt handheld radio power is only 0.2 of the 25-Watt fixed radio.

The antenna used with a fixed radio typically has a gain of 2. The antenna on a handheld has a gain of about 0.5 (or less); the handheld antenna gain compared to the fixed antenna is thus (0.5/2 or) about 0.25.

The antenna height of a handheld is determined by who is holding it, and may be lower than the antenna height of a fixed radio. Assume the handheld is at half the height of the fixed radio antenna. When antenna height is halved, received signal strength decreases by 0.25.

Now we multiply the gain factors that negatively affect the handheld to get the total gain comparison to a fixed-mount radio:

Transmitter Power x Antenna x Height

0.2 x 0.25 x 0.25 = 0.0125

The effective signal of the handheld will then be about 0.0125 of the fixed radio. To express this ratio in decibels:

dB = 10 log(0.0125)

dB = -19 dB

On many typical radio communication paths, a fixed mount radio will have a signal that is very much stronger than necessary and communication is reliable. If a handheld radio is used on that same path, then the signal will be -19 dB (weaker). The handheld radio signal may still be useful, or it may be marginal, or it may be completely uncopied; it depends entirely on the path in use.

The notion that a handheld radio will always be able to communication over the same path as a fixed mount radio with more power, more antenna gain, and a higher antenna is seriously flawed. As demonstrated above, the handheld is at a disadvantage of 0.0125 in total signal power. Or, expressed as the advantage for the fixed mount radio, an advantage of 80:1 in signal power for the fixed mount radio.

To determine how a reduction in signal power by a factor of 0.0125 will affect distance, we must know how path loss changes with distance. Signal power is inversely proportional to distance, and generally we can say that if the distance increases by a factor of 2 the signal strength decreases by a factor of 1/(2^n) where n is at least 2 (for the best possible case) and as high as 4 for typical propagation. Assuming strength decreases by a factor of 1/(2^4) when distance is doubled means at twice the distance the signal is 1/16th as strong. If the fixed mount radio has an signal power advantage of 80, this would be presumed to allow its range to be about three times the the range of the handheld radio.

This means that if a handheld radio signal became uncopyable at a particular distance, then the fixed mount radio signal should be able to be copied farther by a factor of about three-times.

The range of a handheld radio compared to a fixed mount radio and antenna will generally be less. You can estimate the decrease in range by comparing the power output, the antenna gain, and the antenna height.

The power output of a handheld is typically only 5 Watts. A fixed radio can output 25-Watts. A 5-Watt handheld radio power is only 0.2 of the 25-Watt fixed radio.

The antenna used with a fixed radio typically has a gain of 2. The antenna on a handheld has a gain of about 0.5 (or less); the handheld antenna gain compared to the fixed antenna is thus (0.5/2 or) about 0.25.

The antenna height of a handheld is determined by who is holding it, and may be lower than the antenna height of a fixed radio. Assume the handheld is at half the height of the fixed radio antenna. When antenna height is halved, received signal strength decreases by 0.25.

Now we multiply the gain factors that negatively affect the handheld to get the total gain comparison to a fixed-mount radio:

Transmitter Power x Antenna x Height

0.2 x 0.25 x 0.25 = 0.0125

The effective signal of the handheld will then be about 0.0125 of the fixed radio. To express this ratio in decibels:

dB = 10 log(0.0125)

dB = -19 dB

On many typical radio communication paths, a fixed mount radio will have a signal that is very much stronger than necessary and communication is reliable. If a handheld radio is used on that same path, then the signal will be -19 dB (weaker). The handheld radio signal may still be useful, or it may be marginal, or it may be completely uncopied; it depends entirely on the path in use.

The notion that a handheld radio will always be able to communication over the same path as a fixed mount radio with more power, more antenna gain, and a higher antenna is seriously flawed. As demonstrated above, the handheld is at a disadvantage of 0.0125 in total signal power. Or, expressed as the advantage for the fixed mount radio, an advantage of 80:1 in signal power for the fixed mount radio.

To determine how a reduction in signal power by a factor of 0.0125 will affect distance, we must know how path loss changes with distance. Signal power is inversely proportional to distance, and generally we can say that if the distance increases by a factor of 2 the signal strength decreases by a factor of 1/(2^n) where n is at least 2 (for the best possible case) and as high as 4 for typical propagation. Assuming strength decreases by a factor of 1/(2^4) when distance is doubled means at twice the distance the signal is 1/16th as strong. If the fixed mount radio has an signal power advantage of 80, this would be presumed to allow its range to be about three times the the range of the handheld radio.

This means that if a handheld radio signal became uncopyable at a particular distance, then the fixed mount radio signal should be able to be copied farther by a factor of about three-times.

### Re: Communication Range of Handheld Radio

Calculating the Distance for a Certain Power Ratio

Assume at a distance d1 the received power is p1 and at some greater distance d2 the received power is p2.

Now we must state a rate of signal loss with distance. For propagation of VHF signals over typical terrain or sea, a reasonable estimate of rate of signal loss with distance (the path loss) is everytime the distance increases by a factor of 2 the the signal power will decrease by a factor of 1/24. Then, at d=2 the power p would be p=1/24 or 1/16. We express this in a formula where p=power ratio (p2/p1) and d=distance ratio (d2/d1):

p = 1/(d4)

In the context of this discussion of handheld radio range, we want to know the distance ratio when the power ratio will fall to 0.0125. This is ratio of the signal from a handheld radio to a fixed-mount radio. First we rewrite the relationship between p and d to solve for d:

p = 1/(d4)

p × d4 = 1

d4 = 1/p

d = (1/p)0.25

Now we solve for d when p=0.0125 (the power ratio between fixed and handheld signals)

d = (1/0.0125)0.25

d = 800.25

d = 2.98

This tells us that if the rate of signal loss with distance causes the signal to decrease to 1/16 everytime the distance doubles, then a signal will decrease to 0.0125 of its original strength when the distance increases by 2.98 times.

We can also express the rate of attenuation with a logarithm. If the ratio of the power at distance d=1 is p1and at distance d=2 is p2, and that ratio of power was defined to be 1/16, then we can say

p2 / p1 = 0.0625 when distance doubles.

We express this power ratio in deciBels according to the definition of a deciBel:

db = 10 log (p2 / p1)

Evaluating at p2 / p1 = 0.0625 we find

db = 10 log (0.0625)

Now we can describe the signal attenuation at distance in deciBels as a function of the log of the distance and some unknown, x:

dB = x log(d)

Solving for x:

x = dB/log(d)

We solve this for x at distance d=2 and dB = -10 log (0.0625)

x = (-10 log (0.0625))/log(2)

x = -40

Now we can express attenuation in deciBels as a logarithm function of distance as

dB = -40 log(d)

Going back to our original power level ratio between handheld and fixed-mount signals, we found the power ratio in deciBels was -19 dB. We can then solve to find the distance ratio d:

-19 = -40 log(d)

-19/-40 = log(d)

d = 10(-19/-40)

d = 2.98

This is the ratio of the distances when the power will decrease by -19 dB. The distance must increase by a ratio of 2.98:1.

Again, the notion of relating a power decrease ratio to a distance increase ratio is always determined by the assumed rate of decrease with distance. This rate of power decrease with distance is at the best a ratio of 1/d2 and occurs only when the path is in free space, a perfect vacuum with no other conducting bodies anywhere. In real world conditions where the signal propagates in the atmosphere and there are surrounding bodies, terrain, water, obstructions, and so on, the rate of signal power decrease will always be greater than the free space ratio.

Assume at a distance d1 the received power is p1 and at some greater distance d2 the received power is p2.

Now we must state a rate of signal loss with distance. For propagation of VHF signals over typical terrain or sea, a reasonable estimate of rate of signal loss with distance (the path loss) is everytime the distance increases by a factor of 2 the the signal power will decrease by a factor of 1/24. Then, at d=2 the power p would be p=1/24 or 1/16. We express this in a formula where p=power ratio (p2/p1) and d=distance ratio (d2/d1):

p = 1/(d4)

In the context of this discussion of handheld radio range, we want to know the distance ratio when the power ratio will fall to 0.0125. This is ratio of the signal from a handheld radio to a fixed-mount radio. First we rewrite the relationship between p and d to solve for d:

p = 1/(d4)

p × d4 = 1

d4 = 1/p

d = (1/p)0.25

Now we solve for d when p=0.0125 (the power ratio between fixed and handheld signals)

d = (1/0.0125)0.25

d = 800.25

d = 2.98

This tells us that if the rate of signal loss with distance causes the signal to decrease to 1/16 everytime the distance doubles, then a signal will decrease to 0.0125 of its original strength when the distance increases by 2.98 times.

We can also express the rate of attenuation with a logarithm. If the ratio of the power at distance d=1 is p1and at distance d=2 is p2, and that ratio of power was defined to be 1/16, then we can say

p2 / p1 = 0.0625 when distance doubles.

We express this power ratio in deciBels according to the definition of a deciBel:

db = 10 log (p2 / p1)

Evaluating at p2 / p1 = 0.0625 we find

db = 10 log (0.0625)

Now we can describe the signal attenuation at distance in deciBels as a function of the log of the distance and some unknown, x:

dB = x log(d)

Solving for x:

x = dB/log(d)

We solve this for x at distance d=2 and dB = -10 log (0.0625)

x = (-10 log (0.0625))/log(2)

x = -40

Now we can express attenuation in deciBels as a logarithm function of distance as

dB = -40 log(d)

Going back to our original power level ratio between handheld and fixed-mount signals, we found the power ratio in deciBels was -19 dB. We can then solve to find the distance ratio d:

-19 = -40 log(d)

-19/-40 = log(d)

d = 10(-19/-40)

d = 2.98

This is the ratio of the distances when the power will decrease by -19 dB. The distance must increase by a ratio of 2.98:1.

Again, the notion of relating a power decrease ratio to a distance increase ratio is always determined by the assumed rate of decrease with distance. This rate of power decrease with distance is at the best a ratio of 1/d2 and occurs only when the path is in free space, a perfect vacuum with no other conducting bodies anywhere. In real world conditions where the signal propagates in the atmosphere and there are surrounding bodies, terrain, water, obstructions, and so on, the rate of signal power decrease will always be greater than the free space ratio.